Asked by Bradley
Question: What mass of water, in grams, would have to be decomposed to produce 17.6 L of molecular oxygen at STP?
Work:
n=PV/RT
x=(1 atm)(17.6 L)/(0.0821)(273 k)
x = 1.27348295454546
1.27348295454546 x 2 = 2.54696590909092
Grams= Moles x 18.0
Grams = 2.54696590909092 x 18.0
Grams = 45.8
I think I'm messing up somewhere, but I don't know what I'm doing wrong.
Work:
n=PV/RT
x=(1 atm)(17.6 L)/(0.0821)(273 k)
x = 1.27348295454546
1.27348295454546 x 2 = 2.54696590909092
Grams= Moles x 18.0
Grams = 2.54696590909092 x 18.0
Grams = 45.8
I think I'm messing up somewhere, but I don't know what I'm doing wrong.
Answers
Answered by
DrBob222
Your first mistake (I didn't look past that one) is that you didn't plug in the numbers into the calculator. You have the reciprocal of the number of moles.
n = 1*17.6/[(0.09206)(273)]
n should be 0.7857 and that should be rounded to the correct number of significant figures. (I wonder why you didn't just divide 17.6/22.4 to arrive at the number).
n = 1*17.6/[(0.09206)(273)]
n should be 0.7857 and that should be rounded to the correct number of significant figures. (I wonder why you didn't just divide 17.6/22.4 to arrive at the number).